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Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting...

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Singly ionized (one electron removed) atoms are accelerated andthen passed through a velocity selector consisting of perpendicularelectric and magnetic fields. The electric field is 160 V/m and themagnetic field is 3.18×10?2 T . The ions next enter auniform magnetic field of magnitude 1.73×10?2 T that isoriented perpendicular to their velocity.

(A) If the radius of the path of the ions in the second magneticfield is 17.1 cm , what is their mass?

Answer & Explanation Solved by verified expert
4.5 Ratings (978 Votes)

Given E = 160 V/m

          B =3.18×10?2    T

Now the velocity of the ion when it emerges from the velocity selector is

              v = E / B................(1)

                 = (160 V/m) / (3.18×10?2    T)

                = 5031.44 m/s

(a)

magnetic field B =1.75×10?2 T

radius of the path of the ions r = 17.5 cm = 0.175 m

magnetic force F = Bvq   ................... (1)

from Newton's second law of motion ,

      force F = ma  

here , centripetal acceleration a = v2/r

      force F = mv2/r   ................ (2)  

compare eq (1) & eq (2) , we get

         Bvq = mv2/r  

         Bq = mv/r  

mass m = Bqr/v

             = (1.75×10?2 T)(1.6*10-19 C)(0.175 m) /(5031.44 m/s)

             = 9.738*10-26 kg
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